Problem: You have found the following ages (in years) of all 5 seals at your local zoo: $ 5,\enspace 23,\enspace 4,\enspace 5,\enspace 5$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{5 + 23 + 4 + 5 + 5}{{5}} = {8.4\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-3.4$ years $11.56$ years $^2$ $23$ years $14.6$ years $213.16$ years $^2$ $4$ years $-4.4$ years $19.36$ years $^2$ $5$ years $-3.4$ years $11.56$ years $^2$ $5$ years $-3.4$ years $11.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{11.56} + {213.16} + {19.36} + {11.56} + {11.56}} {{5}} $ $ {\sigma^2} = \dfrac{{267.2}}{{5}} = {53.44\text{ years}^2} $ The average seal at the zoo is 8.4 years old. The population variance is 53.44 years $^2$.